Homework 2: Higher Order Functions
Due by 11:59pm on Thursday, July 7
Instructions
Download hw02.zip. Inside the archive, you will find
a file called hw02.py, along with a copy of the ok
autograder.
Submission: When you are done, submit with python3 ok
--submit
. You may submit more than once before the deadline; only the
final submission will be scored. Check that you have successfully submitted
your code on okpy.org. See Lab 0 for more instructions on
submitting assignments.
Using Ok: If you have any questions about using Ok, please refer to this guide.
Readings: You might find the following references useful:
Grading: Homework is graded based on correctness. Each incorrect problem will decrease the total score by one point. There is a homework recovery policy as stated in the syllabus. This homework is out of 2 points.
Required questions
Several doctests refer to these functions:
from operator import add, mul
square = lambda x: x * x
identity = lambda x: x
triple = lambda x: 3 * x
increment = lambda x: x + 1
Getting Started Videos
These videos may provide some helpful direction for tackling the coding problems on this assignment.
To see these videos, you should be logged into your berkeley.edu email.
Q1: Product
Write a function called product
that returns term(1) * ... * term(n)
.
def product(n, term):
"""Return the product of the first n terms in a sequence.
n: a positive integer
term: a function that takes one argument to produce the term
>>> product(3, identity) # 1 * 2 * 3
6
>>> product(5, identity) # 1 * 2 * 3 * 4 * 5
120
>>> product(3, square) # 1^2 * 2^2 * 3^2
36
>>> product(5, square) # 1^2 * 2^2 * 3^2 * 4^2 * 5^2
14400
>>> product(3, increment) # (1+1) * (2+1) * (3+1)
24
>>> product(3, triple) # 1*3 * 2*3 * 3*3
162
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q product
Q2: Accumulate
Let's take a look at how product
is an instance of a more
general function called accumulate
, which we would like to implement:
def accumulate(merger, start, n, term):
"""Return the result of merging the first n terms in a sequence and start.
The terms to be merged are term(1), term(2), ..., term(n). merger is a
two-argument commutative function.
>>> accumulate(add, 0, 5, identity) # 0 + 1 + 2 + 3 + 4 + 5
15
>>> accumulate(add, 11, 5, identity) # 11 + 1 + 2 + 3 + 4 + 5
26
>>> accumulate(add, 11, 0, identity) # 11
11
>>> accumulate(add, 11, 3, square) # 11 + 1^2 + 2^2 + 3^2
25
>>> accumulate(mul, 2, 3, square) # 2 * 1^2 * 2^2 * 3^2
72
>>> # 2 + (1^2 + 1) + (2^2 + 1) + (3^2 + 1)
>>> accumulate(lambda x, y: x + y + 1, 2, 3, square)
19
>>> # ((2 * 1^2 * 2) * 2^2 * 2) * 3^2 * 2
>>> accumulate(lambda x, y: 2 * x * y, 2, 3, square)
576
>>> accumulate(lambda x, y: (x + y) % 17, 19, 20, square)
16
"""
"*** YOUR CODE HERE ***"
accumulate
has the following parameters:
term
andn
: the same parameters as inproduct
merger
: a two-argument function that specifies how the current term is merged with the previously accumulated terms.start
: value at which to start the accumulation.
For example, the result of accumulate(add, 11, 3, square)
is
11 + square(1) + square(2) + square(3) = 25
Note: You may assume that
merger
is commutative. That is,merger(a, b) == merger(b, a)
for alla
,b
, andc
. However, you may not assumemerger
is chosen from a fixed function set and hard-code the solution.
After implementing accumulate
, show how summation
and product
can both be
defined as function calls to accumulate
.
Important:
You should have a single line of code (which should be a return
statement)
in each of your implementations for summation_using_accumulate
and
product_using_accumulate
, which the syntax check will check for.
def summation_using_accumulate(n, term):
"""Returns the sum: term(0) + ... + term(n), using accumulate.
>>> summation_using_accumulate(5, square)
55
>>> summation_using_accumulate(5, triple)
45
>>> # You aren't expected to understand the code of this test.
>>> # Check that the bodies of the functions are just return statements.
>>> # If this errors, make sure you have removed the "***YOUR CODE HERE***".
>>> import inspect, ast
>>> [type(x).__name__ for x in ast.parse(inspect.getsource(summation_using_accumulate)).body[0].body]
['Expr', 'Return']
"""
"*** YOUR CODE HERE ***"
def product_using_accumulate(n, term):
"""Returns the product: term(1) * ... * term(n), using accumulate.
>>> product_using_accumulate(4, square)
576
>>> product_using_accumulate(6, triple)
524880
>>> # You aren't expected to understand the code of this test.
>>> # Check that the bodies of the functions are just return statements.
>>> # If this errors, make sure you have removed the "***YOUR CODE HERE***".
>>> import inspect, ast
>>> [type(x).__name__ for x in ast.parse(inspect.getsource(product_using_accumulate)).body[0].body]
['Expr', 'Return']
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q accumulate
python3 ok -q summation_using_accumulate
python3 ok -q product_using_accumulate
Q3: Filtered Accumulate
Extend the accumulate
function from the previous question to allow for filtering the results
produced by its term
argument by filling in the implementation for the
filtered_accumulate
function:
def filtered_accumulate(merger, start, cond, n, term):
"""Return the result of merging the terms in a sequence of N terms
that satisfy the condition cond. merger is a two-argument function.
If v1, v2, ..., vk are the values in term(1), term(2), ..., term(N)
that satisfy cond, then the result is
start merger v1 merger v2 ... merger vk
(treating merger as if it were a binary operator, like +). The
implementation uses accumulate.
>>> filtered_accumulate(add, 0, lambda x: True, 5, identity) # 0 + 1 + 2 + 3 + 4 + 5
15
>>> filtered_accumulate(add, 11, lambda x: False, 5, identity) # 11
11
>>> filtered_accumulate(add, 0, odd, 5, identity) # 0 + 1 + 3 + 5
9
>>> filtered_accumulate(mul, 1, greater_than_5, 5, square) # 1 * 9 * 16 * 25
3600
>>> # Do not use while/for loops or recursion
>>> from construct_check import check
>>> # ban iteration and recursion
>>> check(HW_SOURCE_FILE, 'filtered_accumulate', ['While', 'For', 'Recursion'])
True
"""
def merge_if(x, y):
"*** YOUR CODE HERE ***"
return accumulate(merge_if, start, n, term)
def odd(x):
return x % 2 == 1
def greater_than_5(x):
return x > 5
filtered_accumulate
has the following parameters:
merger
,start
,n
andterm
: the same arguments asaccumulate
.cond
: a one-argument function applied to the values ofterm(k)
for eachk
from 1 ton
. Only values for whichcond
returns a true value are included in the accumulated total. If no values satisfycond
, thenstart
is returned.
For example, the result of filtered_accumulate(add, 0, is_prime, 11,
identity)
is
0 + 2 + 3 + 5 + 7 + 11
for a valid definition of is_prime
.
Implement filtered_accumulate
by defining the merge_if
function. Exactly
what this function does is something for you to discover. Do not use any loops
or make any recursive calls to filtered_accumulate
.
Hint: The order in which you pass the arguments to
merger
in your solution toaccumulate
matters here.
Use Ok to test your code:
python3 ok -q filtered_accumulate
Q4: Funception
Write a function (funception) that takes in another function func_a
and a number start
and returns a function (func_b
) that will have
one parameter to take in the stop value. func_b
should take the
following into consideration the following in order:
- Takes in the stop value.
- If the value of
start
is less than 0, exit the function by returningNone
. - If the value of
start
is greater than stop, applyfunc_a
onstart
and return the result. - If not, apply
func_a
on all the numbers from start (inclusive) up to stop (exclusive) and return the product.
def funception(func_a, start):
""" Takes in a function (function A) and a start value.
Returns a function (function B) that will find the product of
function A applied to the range of numbers from
start (inclusive) to stop (exclusive)
>>> def func_a(num):
... return num + 1
>>> func_b1 = funception(func_a, 0)
>>> func_b1(3) # func_a(0) * func_a(1) * func_a(2) = 1 * 2 * 3 = 6
6
>>> func_b2 = funception(func_a, 1)
>>> func_b2(4) # func_a(1) * func_a(2) * func_a(3) = 2 * 3 * 4 = 24
24
>>> func_b3 = funception(func_a, 3)
>>> func_b3(2) # Returns func_a(3) since start > stop
4
>>> func_b4 = funception(func_a, -2)
>>> func_b4(-3) # Returns None since start < 0
>>> func_b5 = funception(func_a, -1)
>>> func_b5(4) # Returns None since start < 0
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q funception
Submit
Make sure to submit this assignment by running:
python3 ok --submit
Bonus Questions
Homework assignments will also contain prior exam-level questions for you to take a look at. These questions have no submission component; feel free to attempt them if you'd like a challenge!
Note that exams from Spring 2020, Fall 2020, and Spring 2021 gave students access to an interpreter, so the question format may be different than other years. Regardless, the questions included remain good exam-level problems doable without access to an interpreter.
- Fall 2019 MT1 Q3: You Again [Higher Order Functions]
- Spring 2021 MT1 Q4: Domain on the Range [Higher Order Functions]
- Fall 2021 MT1 Q1b: tik [Functions and Expressions]